Question: What is the value of the following logarithm? $\log_{2} 2$
Explanation: If $b^y = x$ , then $\log_{b} x = y$ Therefore, we want to find the value $y$ such that $2^{y} = 2$ Any number raised to the power $1$ is simply itself, so $2^{1} = 2$ and thus $\log_{2} 2 = 1$.